The other two solutions are also results of generating points from uniform distribution: radius and point on radius result from uniformly drawing pair from [0,

The other two solutions are also results of generating points from uniform distribution: radius and point on radius result from uniformly drawing pair from [0, 2pi) x [0, 1) (radius, distance from center), while "random midpoint" draws from the uniform distribution on the open unit disc. There's no "dependency" at the drawing time. However, if one remembers the substitution formula for multiple variable integrals, this hints why the Jacobian multiplier is necessary in it: all three underlying spaces mentioned ([0,2pi)^2, [0,2pi)x[0,1], B^2\boundary) are diffeomorphic. They clearly don't have the same measure though -- the first one measures 4pi^2, second 2pi, and third pi. However, one can obtain one value from another by integrating Jacobian of appropriate diffeomorphism.